when 68 g of NH3 react with 192 g of O2 then calculate the amount of NO formed (in moles)?

Question

How much NO formed (in moles)

Answer 1

Hello Arshad

 

First let's write the balanced chemical reaction between ammonia and oxygen

 

4NH₃ (g) + 5O₂ (g) → 4NO (g) + 6H₂O (g)

 

Since the masses of NH₃ and O₂ are given, we need to first determine which of them is the limiting reagent.  The limiting reagent will be completely consumed in the reaction, and the amount of product (NO) formed will depend on this value.

 

For NH₃ as the reactant, the stoichiometry process is g NH₃ →  mol NH₃ → mol NO.  Molar mass of NH₃ = 17 g/mol. From the balanced reaction, 4 moles of NH₃ reacts to form 4 moles of NO.

 

68 g NH₃ x (1 mol NH₃ / 17 g NH₃) x (4 mol NO / 4 mol NH₃) = 4.0 mol NO

 

For O₂ as the reactant, the stoichiometry process is g O₂ → mol O₂ → mol NO. Molar mass of O₂ = 32 g/mol. From the balanced reaction, 5 moles of O₂ reacts to form 4 moles of NO.

192 g O₂ x (1 mol O₂ / 32 g O₂) x (4 mol NO / 5 mol O₂) = 4.8 mol NO

 

Since NH₃ forms less NO than O₂, NH₃ is therefore the limiting reagent.  The amount of NO formed in this reaction is 4.0 moles.